# Using the free energy of a monatomic ideal gas, derive

Using the free energy of a monatomic ideal gas, derive the ideal gas law
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toriannucz
The entropy of the system in terms of the partition function and free energy is,
$S=\frac{E}{T}+{k}_{B}{\mathrm{log}}_{e}Z$
Thus, the Helmholtz free energy is,
$F=E-ST$
Substituting the known values,
$F=E-\left(\frac{E}{T}+{k}_{B}{\mathrm{log}}_{e}Z\right)T\phantom{\rule{0ex}{0ex}}=-{k}_{B}T×{\mathrm{log}}_{e}Z$
The partition function for the monoatomic ideal gas is,
$Z={e}^{-N}\left(\frac{V}{N}{\right)}^{N}\left(\frac{2\pi m}{{h}^{2}\beta }{\right)}^{\left(\frac{2}{3}N\right)}$
Substituting the known values,
$F=-{k}_{B}T×{\mathrm{log}}_{e}\left({e}^{-N}\left(\frac{V}{N}{\right)}^{N}\left(\frac{2\pi m}{{h}^{2}\beta }{\right)}^{\left(\frac{3}{2}N\right)}\right)\phantom{\rule{0ex}{0ex}}=-{k}_{B}T×\left({\mathrm{log}}_{e}\left({e}^{-N}\right)+{\mathrm{log}}_{e}\left(\frac{V}{N}{\right)}^{N}+{\mathrm{log}}_{e}\left(\frac{2\pi m}{{h}^{2}\beta }{\right)}^{\frac{3}{2}N}\right)\phantom{\rule{0ex}{0ex}}=-{k}_{B}T×\left(-N+N{\mathrm{log}}_{e}\left(\frac{V}{N}\right)+\left(\frac{3}{2}N\right){\mathrm{log}}_{e}\left(\frac{2\pi m}{{h}^{2}\beta }\right)-1\right)$
As the relation between pressure and the Helmholtz free energy is,
$P=-\left(\frac{dF}{dV}{\right)}_{T,N}\phantom{\rule{0ex}{0ex}}=\frac{{k}_{B}TN}{V}$
As the gas constant is defined as,
$R={N}_{A}{k}_{B}\phantom{\rule{0ex}{0ex}}{k}_{B}=\frac{R}{{N}_{A}}$
where ${N}_{A}$ is an Avogadro's number and ${k}_{B}$ is a Boltzmann constant
Thus, the pressure equation become,
$P=\frac{RTN}{{N}_{A}V}\phantom{\rule{0ex}{0ex}}PV=\frac{N}{{N}_{A}}RT\phantom{\rule{0ex}{0ex}}PV={n}_{a}RT$
where ${n}_{A}$ is the number of moles of gas.
This equation represents the ideal gas equation.